How does one assign measures, using ultraproduct constructions, to the following sets?

Question

If the question is unclear, off topic or has other problems, please explain.

Consider the following sets that are infinitely countable and dense in $\mathbb{R}$

$T_1=\left\{\left.\frac{M_1(n_1)}{R_1(q_1)}\right|n_1,q_1\in\mathbb{Z}\right\}$ $,T_2=\left\{\left.\frac{M_2(n_2)}{R_2(q_2)}\right|n_2,q_2\in\mathbb{Z}\right\},...,$ $T_m=\left\{\left.\frac{M_m(n_m)}{R_m(q_m)}\right|n_m,q_m\in\mathbb{Z}\right\}$

Where $M_1(x),...,M_m(x)$ and $R_1(x),...,R_m(x)$ are unique functions. For example, $M_1(x)=2x+1$, $R_1(x)=2^x$, $M_2(x)=5x+1$, $R_2(x)=\ln(x)$ and so forth.

I want to construct a measure that compares the size of each set to the size of all the sets.

One user suggested

"It is indeed possible to get measures that behave in a nontrivial fashion on countable subsets, and indeed the axiom of choice or some weaker forms thereof are relevant here. Namely, one can get a finitely-additive (not $\sigma$-additive) measure $\xi$ on the set of subsets of $\mathbb{N}$ such that $\xi(S)=0$ for any finite set $S\subseteq\mathbb{N}$ and $\xi(S)=1$ for any cofinite subset (i.e., subset with a finite complement). Here $\xi$ takes only two values, zero or one. For sets that are neither finite nor cofinite, the measure behaves in a nontrivial way but always exactly one of a pair of complementary sets has measure one. Such measures are used in ultraproduct constructions."

I read articles on ultrafilters and ultraproducts, but the mathematics is far beyond my knowledge.

How does one assign measures, that use ultraproduct constructions, to sets $T_1,T_2,..,T_m$?

Answer 1

Firstly, in the formal sense of the word measure, if you can measure all singleton sets $\{x\}$, then the measure of a countable set $T$ is $$\mu(T)=\sum_{x\in T}\mu(\{x\})$$ because measures are defined to be countably additive. That is, you assign a weight to every value, and just sum them up. That's not particularly interesting, especially when one realizes that, if one assigns a positive weight to more than countably many points, then you will end up with countable sets of infinite measure.

However, it seems like you are talking about the situation where you relax countable additivity to finite additivity. In this case, it is true that you can construct "measures" $\mu$ which act non-trivially on countable sets, but the issue is that ultraproduct constructions will not yield any meaningful information because, in such constructions, you are offered infinitely many choices and the measures of your sets will be determined by these choices.

Basically, the way the ultraproduct works is you try to create a set of "big" sets $U$ such that if a set contains a big set, it is itself big and such that, for any subset $X$ either $X$ or its complement is big, but not both. The way you do this is you see that, if you have some set of "big" sets $U$ satisfying the first condition that a set containing a big set is big and that there is no set $X$ such that both $X$ and $\mathbb R\setminus X$ are big. Either $U$ already contains every set or its complete, or there is some set $Y$ such that neither $Y$ nor $\mathbb R\setminus Y$ are yet declared big. You can check that arbitrarily declaring one of these and all sets containing it to be big maintains the conditions we demanded of $U$. Then, Zorn's lemma (which is equivalent to the axiom of choice) tells us, more or less, that we can do this process until we have chosen every set or its complement to be big, leaving us with a set $U$ with the desired properties. You would then declare $\mu(S)=1$ for any big $S$ and $\mu(S)=0$ otherwise.

The issue is that, when you're doing this, unless some of your $T_i$ contain each other, you can just choose whether they're big or small immediately, without relation to anything! So the result of this construction is completely arbitrary. That is to say, you could get any answer you wanted out of this.